User:Frostburn/Collatz scales

The Collatz conjecture states that a certain sequence always reaches one regardless of the starting value.

Definition

We can formulate the Collatz iteration musically as follows:

Start with a seed interval [math]s_0 = \frac{p_0}{q_0}[/math] such that [math]1 \le s_0 \lt 2[/math] and apply the following recursion:

[math] s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 3 p_n}{q_n}\ \mathrm{red}\ 2 [/math]

where [math]\frac{p_n}{q_n}[/math] is in reduced form (no common factors) and [math]x\ \mathrm{red}\ 2[/math] denotes octave reduction i.e. repeated division or multiplication by 2 until [math]1 \le x \lt 2[/math].

We call the set [math]\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 2 \right\}[/math] the Collatz scale of [math]s_0[/math].

It might seem that we need to somehow compute and access the components of the reduced form, but we can make the iteration unconditional by generalizing gcd and lcm to fractions.

Now the recursion is simply. [math] s_{n+1} = \bigl ( (1 + 3\ \mathrm{lcm}(1, s_n)) \cdot \mathrm{gcd}(1, s_n) \bigr ) \mathrm{red}\ 2 [/math]

Conjecture

All Collatz scales are finite.

Evidence

Seed [math]\frac{7}{4}[/math]:

! collatz7.scl
!
Collatz scale of 7/4
6
!
17/16
5/4
11/8
13/8
7/4
2/1

Seed [math]\frac{10}{7}[/math]:

! collatz10_7.scl
!
Collatz scale of 10/7
10
!
35/32
31/28
5/4
71/56
10/7
23/16
53/32
47/28
107/56
2/1

Seed [math]\frac{13}{11}[/math]:

! collatz13_11.scl
!
Collatz scale of 13/11
7
!
23/22
13/11
53/44
61/44
35/22
20/11
2/1

Additional results

Not all variations of the recursion work for all seeds e.g.

[math] s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 5 p_n}{3 q_n}\ \mathrm{red}\ 2 [/math]

Works for [math]\frac{14}{11}[/math]

! varcollatz14_11.scl
!
Collatz scale 5/3 variation of 14/11
7
!
71/66
7/6
14/11
4/3
3/2
178/99
2/1

But produces an infinite scale for [math]\frac{19}{17}[/math].

Additional conjectures

The recursion:

[math] s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 3 p_n}{4 q_n}\ \mathrm{red}\ 5 [/math]

produces finite scales:

[math]\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 5 \right\}[/math]

for all starting values [math]s_0[/math].

Evidence

! varcollatz1_3_4_5_seed_4_3.scl
!
Collatz scale 1, 3, 4, 5 variation of 7/5
13
!
17/16
11/10
11/8
7/5
7/4
2/1
37/16
5/2
49/16
13/4
65/16
17/4
5/1

The recursion:

[math] s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 5 p_n}{4 q_n}\ \mathrm{red}\ 3 [/math]

produces finite scales:

[math]\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 3 \right\}[/math]

for all starting values [math]s_0[/math].

Evidence

! varcollatz1_5_4_3_seed_12_7.scl
!
Collatz scale 1, 5, 4, 3 variation of 12/7
13
!
383/336
7/6
479/336
3/2
12/7
599/336
2/1
61/28
107/48
153/56
11/4
67/24
3/1

There are many (too many to list) other variations of the recursion that seem to work too, but these stood out.